Gazeta Matematica 6-7-8/2016, Problem 27253

Problem:
Evaluate $$\sum_{n=6}^{\infty} \dfrac{n^3-12n^2+47n-60}{n^5-5n^3+4n}$$

Proposed by Alessandro Ventullo, Milan, Italy

Solution:
Observe that $$n^3-12n^2+47n-60=(n-5)(n-4)(n-3)$$ and $$n^5-5n^3+4n=(n-2)(n-1)n(n+1)(n+2),$$ so we have to evaluate
$$\sum_{n=6}^{\infty} \dfrac{(n-5)(n-4)(n-3)}{(n-2)(n-1)n(n+1)(n+2)}.$$
Set $f(x)=\dfrac{(x-5)(x-4)(x-3)}{(x-2)(x-1)x(x+1)(x+2)}$. Since $x=0,\pm 1,\pm 2$ are simple poles, then $$\textrm{Res}(f,2)=\lim_{x \to 2}(x-2)f(x)=-\dfrac{1}{4},$$ $$\textrm{Res}(f,1)=\lim_{x \to 1}(x-1)f(x)=4,$$ $$\textrm{Res}(f,0)=\lim_{x \to 0}xf(x)=-15,$$ $$\textrm{Res}(f,-1)=\lim_{x \to -1}(x+1)f(x)=20,$$ $$\textrm{Res}(f,-2)=\lim_{x \to -2}(x+2)f(x)=-\dfrac{35}{4}.$$
So, $$\begin{array}{lll} \dfrac{(n-5)(n-4)(n-3)}{(n-2)(n-1)n(n+1)(n+2)}&=&-\dfrac{1}{4(n-2)}+\dfrac{4}{n-1}-\dfrac{15}{n}+\dfrac{20}{n+1}-\dfrac{35}{4(n+2)}\\  &=&\left(-\dfrac{1}{4(n-2)}+\dfrac{1}{4(n+2)}\right)+\left(\dfrac{4}{n-1}-\dfrac{4}{n+1}\right)+\\&+&\left(-\dfrac{15}{n}+\dfrac{15}{n+1}\right)+\left(\dfrac{9}{n+1}-\dfrac{9}{n+2}\right). \end{array}$$
Since $$\sum_{n=6}^{\infty} \left(-\dfrac{1}{4(n-2)}+\dfrac{1}{4(n+2)}\right)=\dfrac{1}{4}\left(-\dfrac{1}{4}-\dfrac{1}{5}-\dfrac{1}{6}-\dfrac{1}{7}\right),$$
$$\sum_{n=6}^{\infty} \left(\dfrac{4}{n-1}-\dfrac{4}{n+1}\right)=4\left(\dfrac{1}{5}+\dfrac{1}{6}\right),$$ $$\sum_{n=6}^{\infty} \left(-\dfrac{15}{n}+\dfrac{15}{n+1}\right)=-\dfrac{15}{6},$$ $$\sum_{n=6}^{\infty} \left(\dfrac{9}{n+1}-\dfrac{9}{n+2}\right)=-\dfrac{9}{8},$$ we have $$\sum_{n=6}^{\infty} \dfrac{(n-5)(n-4)(n-3)}{(n-2)(n-1)n(n+1)(n+2)}=\dfrac{1}{4}\left(-\dfrac{1}{4}-\dfrac{1}{5}-\dfrac{1}{6}-\dfrac{1}{7}\right)+4\left(\dfrac{1}{5}+\dfrac{1}{6}\right)-\dfrac{15}{6}-\dfrac{9}{8}=\dfrac{1}{16}.$$